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Tag: 06NB

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Lemma 31.10.8. Let $X \to Y \to Z$ be morphism of schemes. Let $P$ be one of the following properties of morphisms of schemes: flat, locally finite type, locally finite presentation. Assume that $X \to Z$ has $P$ and that $\{X \to Y\}$ can be refined by an fppf covering of $Y$. Then $Y \to Z$ is $P$.

Proof. Let $\mathop{\rm Spec}(C) \subset Z$ be an affine open and let $\mathop{\rm Spec}(B) \subset Y$ be an affine open which maps into $\mathop{\rm Spec}(C)$. The assumption on $X \to Y$ implies we can find a standard affine fppf covering $\{\mathop{\rm Spec}(B_j) \to \mathop{\rm Spec}(B)\}$ and lifts $x_j : \mathop{\rm Spec}(B_j) \to X$. Since $\mathop{\rm Spec}(B_j)$ is quasi-compact we can find finitely many affine opens $\mathop{\rm Spec}(A_i) \subset X$ lying over $\mathop{\rm Spec}(B)$ such that the image of each $x_j$ is contained in the union $\bigcup \mathop{\rm Spec}(A_i)$. Hence after replacing each $\mathop{\rm Spec}(B_j)$ by a standard affine Zariski coverings of itself we may assume we have a standard affine fppf covering $\{\mathop{\rm Spec}(B_i) \to \mathop{\rm Spec}(B)\}$ such that each $\mathop{\rm Spec}(B_i) \to Y$ factors through an affine open $\mathop{\rm Spec}(A_i) \subset X$ lying over $\mathop{\rm Spec}(B)$. In other words, we have ring maps $C \to B \to A_i \to B_i$ for each $i$. Note that we can also consider $$ C \to B \to A = \prod A_i \to B' = \prod B_i $$ and that the ring map $B \to \prod B_i$ is faithfully flat and of finite presentation.

The case $P = flat$. In this case we know that $C \to A$ is flat and we have to prove that $C \to B$ is flat. Suppose that $N \to N' \to N''$ is an exact sequence of $C$-modules. We want to show that $N \otimes_C B \to N' \otimes_C B \to N'' \otimes_C B$ is exact. Let $H$ be its cohomology and let $H'$ be the cohomology of $N \otimes_C B' \to N' \otimes_C B' \to N'' \otimes_C B'$. As $B \to B'$ is flat we know that $H' = H \otimes_B B'$. On the other hand $N \otimes_C A \to N' \otimes_C A \to N'' \otimes_C A$ is exact hence has zero cohomology. Hence the map $H \to H'$ is zero (as it factors through the zero module). Thus $H' = 0$. As $B \to B'$ is faithfully flat we conclude that $H = 0$ as desired.

The case $P = locally finite type$. In this case we know that $C \to A$ is of finite type and we have to prove that $C \to B$ is of finite type. Because $B \to B'$ is of finite presentation (hence of finite type) we see that $A \to B'$ is of finite type, see Algebra, Lemma 9.6.2. Therefore $C \to B'$ is of finite type and we conclude by Lemma 31.10.2.

The case $P = locally finite presentation$. In this case we know that $C \to A$ is of finite presentation and we have to prove that $C \to B$ is of finite presentation. Because $B \to B'$ is of finite presentation and $B \to A$ of finite type we see that $A \to B'$ is of finite presentation, see Algebra, Lemma 9.6.2. Therefore $C \to B'$ is of finite presentation and we conclude by Lemma 31.10.1. $\square$

\begin{lemma}
\label{lemma-curiosity}
Let $X \to Y \to Z$ be morphism of schemes.
Let $P$ be one of the following properties of morphisms of schemes:
flat, locally finite type, locally finite presentation.
Assume that $X \to Z$ has $P$ and that $\{X \to Y\}$
can be refined by an fppf covering of $Y$. Then $Y \to Z$ is $P$.
\end{lemma}

\begin{proof}
Let $\Spec(C) \subset Z$ be an affine open and let
$\Spec(B) \subset Y$ be an affine open which maps into
$\Spec(C)$. The assumption on $X \to Y$ implies we can
find a standard affine fppf covering $\{\Spec(B_j) \to \Spec(B)\}$
and lifts $x_j : \Spec(B_j) \to X$. Since $\Spec(B_j)$
is quasi-compact we can find finitely many affine opens
$\Spec(A_i) \subset X$ lying over $\Spec(B)$
such that the image of each $x_j$
is contained in the union $\bigcup \Spec(A_i)$. Hence after
replacing each $\Spec(B_j)$ by a standard affine Zariski coverings
of itself we may assume we have a
standard affine fppf covering $\{\Spec(B_i) \to \Spec(B)\}$
such that each $\Spec(B_i) \to Y$ factors through an affine
open $\Spec(A_i) \subset X$ lying over $\Spec(B)$.
In other words, we have ring maps $C \to B \to A_i \to B_i$ for each $i$.
Note that we can also consider
$$
C \to B \to A = \prod A_i \to B' = \prod B_i
$$
and that the ring map $B \to \prod B_i$ is faithfully flat and
of finite presentation.

\medskip\noindent
The case $P = flat$. In this case we know that $C \to A$ is flat
and we have to prove that $C \to B$ is flat. Suppose that
$N \to N' \to N''$ is an exact sequence of $C$-modules. We want to
show that $N \otimes_C B \to N' \otimes_C B \to N'' \otimes_C B$
is exact. Let $H$ be its cohomology and let $H'$ be the cohomology
of $N \otimes_C B' \to N' \otimes_C B' \to N'' \otimes_C B'$. As
$B \to B'$ is flat we know that $H' = H \otimes_B B'$. On the other hand
$N \otimes_C A \to N' \otimes_C A \to N'' \otimes_C A$
is exact hence has zero cohomology. Hence the map
$H \to H'$ is zero (as it factors through the zero module).
Thus $H' = 0$. As $B \to B'$ is faithfully flat we conclude that
$H = 0$ as desired.

\medskip\noindent
The case $P = locally\ finite\ type$.
In this case we know that $C \to A$ is of finite type and
we have to prove that $C \to B$ is of finite type.
Because $B \to B'$ is of finite presentation (hence of finite type)
we see that $A \to B'$ is of finite type, see
Algebra, Lemma \ref{algebra-lemma-compose-finite-type}.
Therefore $C \to B'$ is of finite type and we conclude by
Lemma \ref{lemma-finite-type-local-source-fppf-algebra}.

\medskip\noindent
The case $P = locally\ finite\ presentation$.
In this case we know that $C \to A$ is of finite presentation and
we have to prove that $C \to B$ is of finite presentation.
Because $B \to B'$ is of finite presentation and $B \to A$
of finite type we see that $A \to B'$ is of finite presentation, see
Algebra, Lemma \ref{algebra-lemma-compose-finite-type}.
Therefore $C \to B'$ is of finite presentation and we conclude by
Lemma \ref{lemma-flat-finitely-presented-permanence-algebra}.
\end{proof}

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